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Suppose that $\Sigma_{k=1}^\infty a_k$ converges. Prove that if $b_k\uparrow \infty$ and $\Sigma_{k = 1}^\infty a_kb_k$ converges, then

$b_m \Sigma_{k = m}^\infty a_k → 0$ as $m → \infty$.

Attemtp: Suppose $\Sigma_{k=1}^\infty a_k$ converges, and $\Sigma_{k = 1}^\infty a_kb_k$ also.Then we know by Abel's Formula that the sequences converge only iff its partial sums converge. If we let $\Sigma a_k = \Sigma \frac{a_kb_k}{b_k}$, then because $b_k$ is increasing we have $\frac{1}{b_k}$ is decreasing to zero. So we almost have a telescoping series.

Then let $c_k = \Sigma_{ j = k}^{\infty} a_jb_j$. Then $\Sigma_{k = n}^m a_k = \Sigma \frac{a_kb_k}{b_k}= \Sigma_{k = n}^m \frac{c_k - c_{k+1}}{b_k}$

I don't know how to continue. I am having trouble applying Abel's Formula and the subscripts are confusing. Can someone please help me? I am suppose to use Abel's Formula. Thank you very much.

Abel's Formula: Let $a_k,b_k$ be real sequences, and for each pair of integers $n \geq m \geq 1$ set $A_{n,m} = \Sigma_{k =m}^n a_k$ $\Sigma_{k = n}^m a_kb_k = A_{n,m}b_n - \Sigma_{k = m}^{n-1} A_{k,m}(b_{k+1} -b_k)$

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  • $\begingroup$ Can anyone please help? I am desperate. I need help. Thank you. $\endgroup$ – user 48284 Feb 25 '15 at 2:43
  • $\begingroup$ we know $\sum_{k = m}^\infty a_kb_k\to 0$ and $b_m\sum_{k = m}^\infty a_k\leq\sum_{k = m}^\infty a_kb_k$. We also know $b_k$ is positive after a term $m_0$. The proof would have come if $a_k$ had been also positive after some term, say $m_0$, by comprison test for series. $\endgroup$ – guest Feb 25 '15 at 9:51
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At first I assume that $$ a_{k},b_{k}\ge 0$$ We have 2 convergence based on input data:

$$ \sum_{k=1}^\infty a_{k}=C_{1} , \sum_{k=1}^\infty a_{k}b_{k}=C_{2} $$ we also assume that $$ \lim_{k=\infty} b_{k}=\infty \Rightarrow $$ so for $$ m \rightarrow \infty : b_{m}<b_{m+1}<b_{m+2}<... $$

So by simplifying mentioned equations: $$ \sum_{k=1}^{m-1} a_{k}=D_{1} \Rightarrow D_{1}+\sum_{k=m}^\infty a_{k}=C_{1} $$ $$ \sum_{k=1}^{m-1} a_{k}b_{k}=D_{2} \Rightarrow D_{2}+\sum_{k=m}^\infty a_{k}b_{k}=C_{2} $$ And we know that $$ C_{1},C_{2},D_{1},D_{2}<|M|<\infty$$

In conclusion I could say that : $$ C_{2}-D_{2} =\sum_{k=m}^\infty a_{k}b_{k}=a_{m}b_{m}+a_{m+1}b_{m+1}+a_{m+2}b_{m+2}+...>b_{m}(a_{m}+a_{m+1}+a_{m+2}+...)$$ $$\Rightarrow \infty>|M_{2}|>C_{2}-D_{2}>b_{m}\sum_{k=m}^\infty a_{k}=b_{m}(C_{1}-D_{1})$$ $$\Rightarrow C_{2}-D_{2}>b_{m}(C_{1}-D_{1})$$ By having $$|M_{1}|>C_{1}-D_{1} , \lim_{k=\infty} b_{k}=\infty $$ I can receive to these: $$ C_{1}-D_{1}=C_{2}-D_{2}=0 \Rightarrow \lim_{m=\infty}b_{m}\sum_{k=m}^\infty a_{k}=0$$ $$ $$

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Let $c_k = a_kb_k$, $C_k =\sum_{j\ge k}c_j$. We can observe that $$\begin{align*} \sum_{k\ge m}a_k &=\sum_{k\ge m}\frac{c_k}{b_k}\\ &=\sum_{k\ge m}\frac{C_k-C_{k+1}}{b_k}\\ &=\lim_{N\to\infty}\sum_{k= m}^N\frac{C_k-C_{k+1}}{b_k}\\ &=\lim_{N\to\infty}\left(\sum_{k= m}^N\frac{C_k}{b_k}-\sum_{k= m+1}^{N+1}\frac{C_k}{b_{k-1}}\right)\\ &=\lim_{N\to\infty}\left(\sum_{k= m+1}^NC_k\left(\frac1{b_k}-\frac1{b_{k-1}}\right)+\frac{C_m}{b_m}-\frac{C_{N+1}}{b_N}\right)\\ &=\sum_{k= m+1}^\infty C_k\left(\frac1{b_k}-\frac1{b_{k-1}}\right)+\frac{C_m}{b_m} \end{align*}$$ since $C_k\xrightarrow{k\to\infty} 0$ and $0\le b_k\le b_{k+1}\xrightarrow{k\to\infty} \infty \ \ (k\ge k_0)$. Then, we find that for $m\ge k_0$, $$\begin{align*} \left|b_m\sum_{k\ge m}a_k\right| &=\left|b_m\sum_{k= m+1}^\infty C_k\left(\frac1{b_k}-\frac1{b_{k-1}}\right)\right|+\left|C_m\right|\\ &\le b_m\cdot \sup_{k\ge m+1}|C_k|\cdot \sum_{k\ge m+1} \left(\frac1{b_{k-1}}-\frac1{b_{k}}\right) +|C_m|\\ &\le \sup_{k\ge m+1}|C_k|+|C_m| \xrightarrow{m\to\infty} 0 \end{align*}$$ as wanted.

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